On Certain Class of Meromorphically Multivalent Reciprocal Starlike Functions Associated with the Liu-Srivastava Operator Defined by Subordination

Filomat 8:9 014, 1965 198 DOI 10.98/FIL1409965M Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.rs/filomat On Certain Class of Meromorphically Multivalent Reciprocal Starlike Functions Associated with the Liu-Srivastava Operator Defined by Subordination Li-Na Ma a, Shu-Hai Li a a School of Mathematics and Statistics, Chifeng University, Chifeng 04000, Inner Mongolia, China Abstract. In the paper, we introduce the class of meromorphically p-valent reciprocal starlike functions associated with the Liu-Srivastava operator defined by subordination. Some sufficient conditions for functions belonging to this class are derived. The results presented here improve and generalize some known results. 1. Introduction and Preliminaries Let Σ p denote the class of meromorphic functions of the form f z z p + a k z k p+1 p N {1,, }, 1 which are analytic and p-valent in the punctured open unit disk U {z C : 0 < z < 1} U\{0}, where U is the open unit disk U {z C : z < 1}. In particular, we set Σ 1 Σ. For two functions f and, analytic in U, we say that the function f is subordinate to in U, if there exists a Schwartz function ω, which is analytic in U with such that ω0 0 and ωz < 1 z U, f z ωz z U. We denote this subordination by f z z. Furthermore, if the function is univalent in U, then the following equivalent relationship holdssee for details [4, 11]; see also [17]: f z z f 0 0 and f U U. 010 Mathematics Subject Classification. Primary 30C45; Secondary 30C55 Keywords. Meromorphically multivalent functions; Carlson-Shafer operator; Dziok-Srivastava operator; Liu-Srivastava operator; Starlike functions; Reciprocal; subordination. Received: 03 September 013; Accepted: 4 January 014 Communicated by Hari M. Srivastava Research supported by the Natural Science Foundation of Inner Mongolia under Grant 014MS0101. Email addresses: malina00@163.com Li-Na Ma, lishms66@sina.com Shu-Hai Li

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1966 A function f Σ p is said to be in the class S pα of meromorphically p-valent starlike of order α if it satisfies the inequality z f z R < α 0 α < 1; z U. p f z As usual, we let S p0 S p. Furthermore, a function f S p is said to be in the class M p α of meromorphically p-valent starlike of reciprocal order α if and only if p f z R z f < α 0 α < 1; z U. 3 z In particular, we set M 1 α Mα. Remark 1.1. In view of the fact that Rpz < 0 R 1 pz R < 0, pz pz it follows that a meromorphically p-valent starlike function of reciprocal order 0 is same as a meromorphically p-valent starlike function. When 0 < α < 1, the function f Σ p is meromorphically p-valent starlike of reciprocal order α if and only if z f z p f z + 1 α < 1 z U. α 4 For p 1, this class Mα was considered by Sun et al. [18]. For arbitrary fixed real numbers A and B 1 B < A 1, we denote by PA, B the class of functions of the form qz 1 + c 1 z + c z +, which is analytic in the unit disk U and satisfies the condition qz 1 + Az 1 + Bz z U, where the symbol stands for usual subordination. The class PA, B was introduced and studied by Janowski [8]. We also observe from 5 see, also [15] that a function qz PA, B if and only if 1 AB qz 1 B < A B B 1; z U 6 1 B and R{qz} > 1 A B 1; z U. 7 For functions f Σ p given by 1 and Σ p given by z z p + b k z k p+1 p N, 8 we define the Hadamard product or convolution of f and by f z : z p + a k b k z k p+1 f z. 9 5

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1967 The linear operator L p a, c is defined as followssee[10] L p a, c f z : φ p a, c; z f z f Σ p, 10 and φ p a, c; z is defined by φ p a, c; z : z p + a k z k p+1 z U ; a R; c R \ Z 0 c ; Z 0 0, 1,,, 11 k where λ n is the Pochhammer symbolor the shifted factorial defined in terms of the Gamma function by Γλ + n 1, n 0, λ n 1 Γλ λλ + 1 λ + n 1, n N. The Liu-Srivastava operator L p a, c, analogous to the Carlson-Shafer operator, was considered by Liu and Srivastava [10] on the space of analytic and meromorphically p-valent functions. The carlson-shafer operator L p a, c was introduced earlier by Saitoh [14] on the space of analytic and p-valent functions in U. In present, the Carlson-Shafer operator was generalized to the Dziok-Srivastava operator by Dziok and Srivastava [5, 6]. Recently, Aouf et al.[1] constructed a new operator by applying the Liu-Srivastava operator L p a, c. In [10], making use of the Liu-Srivastava operator L p a, c, Liu and Srivastava discussed the subclass of Σ p such that zl p a, c f z pl p a, c f z 1 + Az 1 + Bz. 13 By using the Liu-Srivastava operator L p a, c, we now introduce a new subclass M a,c p; β; A, B satisfying the following subordination condition for f Σ p given by 1, { } p Lp a, c f z 1 pβ zl p a, c f z + β 1 + Az 1 + Bz, 14 where 1 B < A 1, a > 0, c > 0, 0 pβ < 1, p N. We note that 14 is equivalent to by 6 and 7 { } p Lp a, c f z 1 pβ zl p a, c f z + β + 1 AB 1 B < A B 1 B B 1; z U 15 and R { { }} p Lp a, c f z 1 pβ zl p a, c f z + β < 1 A B 1; z U. 16 We note 16 is equivalent to 1 pβ zl p a, c f z p L p a, c f z + βzl p a, c f z + 1 1 A < 1 1 A B 1, A 1; z U 17 and p L p a, c f z 1 pβ zl p a, c f z + β + 1 < 1 B 1, A 1; z U 18 By specializing the parameters p, a, c, A, B and β, we obtain the following classes studied by different authors,

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1968 1 M 1,1 1; 0; 1 α, 1 Mα0 α < 1 see [18]; M 1,1 1; 0; B, ba B + B Σ [b; A, B]b < 0, 1 B < A 1 see [3]; A Bp a 3 M 1,1 p; 0; B, B + Q k [p, a, A, B]0 a < p, 1 A < B 1, 0 B 1, A + B 0 see p [16]. In recent years, more and more researchers are interested in the reciprocal case of the starlike functionssee [19],[9],[13],[]. In the present investigation, we give some sufficient conditions for the function belonging to the class M a,c p; β; A, B. In order to establish our main results, we need the following lemmas. Lemma 1.. Jack s lemma [7] Let the nonconstant function ωz be analytic in U with ω0 0. If ωz attains its maximum value on the circle z r < 1 at a point z 0 U, then z 0 ω z 0 γωz 0, where γ is a real number and γ 1. Lemma 1.3. [1] Let Ω be a set in the complex plane C and suppose that Φ is a mapping from C U to C which satisfies Φix, y; z Ω for z U, and for all real x, y such that y 1 + x. If the function pz 1+c 1 z+c z + is analytic in U and Φpz, zp z; z Ω for all z U, then Rpz > 0. Lemma 1.4. [0] Let ρz 1 + b 1 z + b z + be analytic in U and η be analytic and starlike with respect to the origin univalent in U with η0 0. If zρ z ηz, then z ρz 1 + 0 ηt dt. t. Main Results Unless otherwise mentioned we shall assume through this paper that a > 0, c > 0, 1 B < A 1, 0 pβ < 1, p N. We begin by presenting the following coefficient sufficient condition for functions belonging to the class M a,c p; β; A, B. Theorem.1. If f Σ p satisfies anyone of the following conditions: i For B 1, k p + 1 + p1 B [1 + k p + 1β] + 1 AB1 pβk p + 1 ak < p1 B, 19 1 pβa B c k ii For B 1, A 1, 1 A1 pβk p + 1 1 + k p + 1β + ak 1 + k p + 1β + p < 1 pβ1 A, 0 c k iii For B 1, A 1, k p + 1 + k + 1 1 pβ then f M a,c p; β; A, B. ak c k < p, 1

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1969 Proof. i If B 1, by the condition 15, we only need to show that p1 B { } Lp a, c f z 1 pβa B zl p a, c f z + β + 1 AB < 1 z U. A B We first observe that p1 B { } Lp a, c f z 1 pβa B zl p a, c f z + β Bp + p + k p + 1 a k c k a k z k+1 < B p + B p + + 1 AB A B p1 B [1+k p+1β]+1 AB1 pβk p+1 a k 1 pβa B c k a k z k+1 p1 B [1+k p+1β]+1 AB1 pβk p+1 a k 1 pβa B c k z k+1 p k p + 1 a k c k z k+1 p1 B [1+k p+1β]+1 AB1 pβk p+1 a k 1 pβa B c k p k p + 1 a k c k Now, by using the inequality 19, we have B p + p1 B [1+k p+1β]+1 AB1 pβk p+1 a k 1 pβa B c k p k p + 1 a k c k 3 < 1, 4 which, in conjunction with 3, completes the proof of i for Theorem.1. ii If B 1, A 1, by virtue of the condition 17, we only need to show that 1 A1 pβ zl p a, c f z p L p a, c f z + βzl p a, c f z + 1 < 1 z U. 5 We first observe that 1 A1 pβ zl p a, c f z p L p a, c f z + βzl p a, c f z + 1 A1 pβ + [ ] 1 A1 pβk p+1 ak 1 + k p + 1β + p c k a k z k+1 1 pβ + [1 + k p + 1β] a k c k a k z k+1 A 1 pβ + 1 + k p + 1β + 1 A1 pβk p+1 p a k c k z k+1 1 pβ 1 + k p + 1β a k c k z k+1 A 1 pβ + 1 + k p + 1β + 1 A1 pβk p+1 p a k c k < 1 pβ 1 + k p + 1β a k c k 6

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1970 Now, by using the inequality 0, we have A1 pβ + 1 + k p + 1β + 1 A1 pβk p+1 p a k c k 1 pβ < 1, 7 1 + k p + 1β a k c k which, in conjunction with 6, completes the proof of ii for Theorem.1. iii If B 1, A 1, by virtue of the condition 18, we only need to show that p L p a, c f z 1 pβ zl p a, c f z + β + 1 < 1 z U. 8 We first observe that p L p a, c f z 1 pβ zl p a, c f z + β + 1 Now, by using the inequality 1, we have p k+1 a k 1 pβ c k k p + 1 a k c k < 1, < k+1 ak 1 pβ c k a k z k+1 p + k p + 1 a k c k a k z k+1 k+1 a k 1 pβ c k z k+1 p p k p + 1 a k c k z k+1 k+1 a k 1 pβ c k k p + 1 a k c k which, in conjunction with 9, completes the proof of iii for Theorem.1. Taking a c 1, β 0 in Theorem.1, we obtain the following Corollary. Corollary.. If f Σ p satisfies anyone of the following conditions: i For B 1, k p + 1 + p1 B + k p + 11 AB < p1 B, A B ii For B 1, A 1, 1 Ak p + 1 1 + 1 + p < 1 A, iii For B 1, A 1, k p + 1 + k + 1 ak < p, then f M 1,1 p; 0; A, B. 9 30

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1971 Taking p 1, A 1 α, 0 < α < 1, B 1 in Corollary., we obtain the following result. Corollary.3. [18] If f Σ satisfies 1 + kα < 1 1 1 α, then f Mα, for 0 < α < 1. Theorem.4. If f Σ p satisfies anyone of the following conditions: i For B 1, 1 + zl pa, c f z L p a, c f z zl pa, c f z L p a, c f z < 1 pβa B 1 pβa B + 1 + B, 31 ii For B 1, 1 < A 0, 1 + zl pa, c f z L p a, c f z zl pa, c f z L p a, c f z < 1 pβ1 A1 + A pβ1 + A + 1 A, 3 iii For B 1, A 1, 1 + zl pa, c f z L p a, c f z zl pa, c f z L p a, c f z < 1 pβ pβ, 33 then f z M a,c p; β; A, B. Proof. i If B 1, let 1 + 1+ B 1+ B +A B p Lp a,c f z 1 pβ zl p a,c f z + β ωz 1 1+ B 1+ B +A B 1 z U. 34 Then the function ω is analytic in U with ω0 0. We easily find from 34 that pl p a, c f z 1 pβa Bωz 1 + B zl p a, c f z 1 + B z U. 35 Differentiating both sides of 35 logarithmically, we obtain 1 + zl pa, c f z L p a, c f z zl pa, c f z L p a, c f z by virtue of 31 and 36, we find that 1 + zl pa, c f z L p a, c f z zl pa, c f z L p a, c f z 1 pβa Bzω z 1 pβa Bωz 1 + B, 36 zω 1 pβa B z 1 pβa Bωz 1 + B < 1 pβa B 1 pβa B + 1 + B. Next, we claim that ωz < 1. Indeed, if not, there exists a point z 0 U such that max ωz ωz 0 1 z 0 U. 37 z z 0

Applying Jack s Lemma to ωz at the point z 0, we have L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 197 z 0 ω z 0 γωz 0 γ 1. Now, upon setting ωz 0 e iθ 0 θ < π. If we put z z 0 in 36, we get 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 z 1 pβa B 0 ω z 0 L p a, c f z 0 1 pβa Bωz 0 1 + B 1 pβa B γ 1 pβa B 1 + B e iθ 1 pβa B 1 1 pβa B 1 + B e iθ. This implies that 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 L p a, c f z 0 [1 pβa B] [1 pβa B] + 1 + B 1 pβa B1 + B cos θ Since the right hand side of 38 takes it minimum value for cos θ 1, we have that 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 [1 pβa B] L p a, c f z 0 [1 pβa B + 1 + B ] 38 This contradicts our condition 31 of Theorem.4. Therefore, we conclude that ωz < 1, which shows that 1 + 1+ B 1+ B +A B p Lp a,c f z 1 pβ zl p a,c f z + β 1 1 1+ B < 1 B 1; z U. 1+ B +A B This implies that { } p Lp a, c f z 1 pβ zl p a, c f z + β + 1 < A B 1 + B B 1; z U, 39 then, we have { } p Lp a, c f z 1 pβ zl p a, c f z + β + 1 AB 1 B p 1 pβ { Lp a, c f z } zl p a, c f z + β < A B B A B + 1 + B 1 B A B B 1; z U. 1 B Therefore, we conclude that f z M a,c p; β; A, B for B 1. ii If B 1, 1 < A 0, analogously to Theorem. in [18], we let ωz 1 + 1 A p 1 pβ 1 Lpa,c f z zlpa,c f z +β 1 1 A + 1 + 1 AB 1 B 1 1 40

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1973 Then the function ω is analytic in U with ω0 0. We easily find from 40 that zl p a, c f z pl p a, c f z 1 A + 1 + Aωz 1 A pβ1 + Aωz z U. 41 Differentiating both sides of 41 logarithmically, we obtain 1 + zl pa, c f z L p a, c f z zl pa, c f z 1 pβ1 A1 + Azω z L p a, c f z 1 A + 1 + Aωz1 A pβ1 + Aωz, 4 by virtue of 3 and 4, we find that 1 + zl pa, c f z L p a, c f z zl pa, c f z L p a, c f z zω 1 pβ1 A1 + A z 1 A + 1 + Aωz1 A pβ1 + Aωz 1 pβ1 A1 + A < pβ1 + A + 1 A. Next, we claim that ωz < 1. Indeed, if not, there exists a point z 0 U such that max ωz ωz 0 1 z 0 U. 43 z z 0 Applying Jack s Lemma to ωz at the point z 0, we have z 0 ω z 0 γωz 0 γ 1. 44 Now, upon setting ωz 0 e iθ 0 θ < π. If we put z z 0 in 4, we get 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 L p a, c f z 0 1 pβ1 A1 + A γ 1 + A 1 Ae iθ pβ1 + A + 1 Ae iθ 1 pβ1 A1 + A 1 pβ1 + A + 1 + pβ1 A1 + Ae iθ 1 A e iθ. This implies that 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 L p a, c f z 0 1 pβ1 A1 + A P cos θ + Q cos θ + R, 45 where P 4pβ1 + A 1 A, Q 1 + pβ1 A1 + A[1 A + pβ1 + A ], R 1 + A [p β 1 + A + 1 A ]. Since the right hand side of 54 takes it minimum value for cos θ 1, we have that 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 L p a, c f z 0 [1 pβ1 A1 + A] [pβ1 + A + 1 A]. 46

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1974 This contradicts our condition 3 of Theorem.4. Therefore, we conclude that ωz < 1, which shows that 1 + 1 A 1 p Lpa,c f z 1 pβ zlpa,c f z +β 1 1 1 A < 1. 47 This implies that 1 pβ p zl p a, c f z L p a, c f z + βzl p a, c f z + 1 < 1, 48 1 A then, we have 1 pβ zl p a, c f z p L p a, c f z + βzl p a, c f z + 1 1 A < 1 pβ p zl p a, c f z L p a, c f z + βzl p a, c f z + 1 + 1 1 A 1 1 A 1 + 1 1 1 A 1 1 < A 0; z U. 1 A Therefore, we conclude that f z M a,c p; β; A, B for B 1, 1 < A 0. iii If B 1, A 1, we let p L p a, c f z ωz 1 pβ zl p a, c f z + β + 1. 49 Then the function ω is analytic in U with ω0 0. We easily find from 49 that pl p a, c f z 1 pβωz 1 z U. 50 zl p a, c f z Differentiating both sides of 50 logarithmically, we obtain 1 + zl pa, c f z L p a, c f z zl pa, c f z 1 pβzω z L p a, c f z 1 pβωz 1, 51 by virtue of 33 and 51, we find that 1 + zl pa, c f z L p a, c f z zl pa, c f z L p a, c f z zω 1 pβ z 1 pβωz 1 < 1 pβ pβ. Next, we claim that ωz < 1. Indeed, if not, there exists a point z 0 U such that max ωz ωz 0 1 z 0 U. 5 z z 0 Applying Jack s Lemma to ωz at the point z 0, we have z 0 ω z 0 γωz 0 γ 1. 53 Now, upon setting ωz 0 e iθ 0 θ < π.

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1975 If we put z z 0 in 51, we get 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 L p a, c f z 0 z 1 pβ 0 ω z 0 1 pβωz 0 1 1 pβ γ 1 pβ e iθ 1 pβ 1 1 pβ e iθ. This implies that 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 L p a, c f z 0 1 pβ 1 + 1 pβ 1 pβ cos θ. 54 Since the right hand side of 54 takes it minimum value for cos θ 1, we have that 1 + z 0L p a, c f z 0 L p a, c f z 0 z 0L p a, c f z 0 L p a, c f z 0 1 pβ pβ 55 This contradicts our condition 33 of Theorem.4. Therefore, we conclude that ωz < 1, which shows that p L p a, c f z 1 pβ zl p a, c f z + β + 1 < 1. 56 This implies that p L p a, c f z 1 pβ zl p a, c f z + β 1 + z 1 z. 57 Therefore, we conclude that f z M a,c p; β; A, B for B 1, A 1. Putting p 1, a c 1, A 1 α, 1 α < 1, B 1 and β 0 in Theorem.4, we obtain the following Corollary. Corollary.5. [18] If f Σ satisfies 1 + z f z f z z f z f z then f Mα, for 1 α < 1. < 1 α, Theorem.6. If f Σ p satisfies R 1 + zl pa, c f z L p a, c f z zl pa, c f z < L p a, c f z then f z M a,c p; β; A, B. 1 A + pβa B 1 pβa B 1 pβa B [1 A + pβa B] B + 1 B 1 pβ A 1 B < A B + 1 B 1 pβ, 58

L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1976 Proof. Suppose that z : { } p Lp a,c f z 1 pβ zl p a,c f z + β 1 A 1 B 1 1 A 1 B 1 B < A 1; z U. 59 Then is analytic in U. It follows from 59 that pl p a, c f z 1 pβa B z + 1 A + pβa B. 60 zl p a, c f z 1 B Differentiating 60 logarithmically, we obtain 1 zl pa, c f z L p a, c f z + zl pa, c f z 1 pβa Bz z L p a, c f z 1 pβa B z + 1 A + pβa B : Φ z, z z; z, where Φr, s; t 1 pβa Bs 1 pβa Br + 1 A + pβa B. For all real x and y satisfying y 1 + x, we have RΦix, y; z 1 pβa By R i1 pβa Bx + 1 A + pβa B 1 pβa B[1 A + pβa B]y [1 A + pβa B] + [1 pβa B] x 1 + x 1 pβa B[1 A + pβa B] [1 A + pβa B] + [1 pβa B] x 1 A + pβa B B + 1 B 1 pβa B 1 pβ A 1, 1 pβa B B < A B + 1 B [1 A + pβa B] 1 pβ. We now put Ω ξ : Rξ > 1 A + pβa B 1 pβa B 1 pβa B [1 A + pβa B] B + 1 B 1 pβ A 1, B < A B + 1 B, 1 pβ then Φix; y; z Ω for all real x, y such that y 1+x. Moreover, in view of 58, we know that Φ z, z z; z Ω. Thus, by Lemma 1.3, we deduce that R z > 0 z U, which shows that the desired assertion of Theorem.6 holds. Putting p 1, a c 1, A 1 α, 0 α < 1, B 1 and β 0 in Theorem.6, we obtain the following Corollary.

Corollary.7. [18] If f Σ satisfies R 1 + z f z f z f α z 1 α < z f z 1 α α then f Mα, for 0 α < 1. L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1977 0 α 1, 1 α < 1, Theorem.8. If f Σ p satisfies { plp a, c f z R 1 zl p a, c f z + η zl pa, c f z } L p a, c f z > 1 then f M a,c p; β; A, B for η 0. Proof. We define the function hz by { } p Lp a,c f z 1 pβ zl p a,c f z + β hz : 1 1 A 1 B 1 A 1 B Then h is analytic in U. It follows from 6 that and where 1 pβa B 1 B pl p a, c f z 1 pβa Bhz + 1 A + pβa B zl p a, c f z 1 B A + pβa B] η + pη 1 η[1, 61 1 B 1 B < A 1; z U. 6 1 + η zl pa, c f z P + Qhz + Rzh z L p a, c f z 1 pβa Bhz + 1 A + pβa B, 64 P pη1 B + 1 η[1 A + pβa B]; Q 1 pβa B1 η; R 1 pβa Bη. Combining 63 and 64, we get pl p a, c f z 1 + η zl pa, c f z where zl p a, c f z L p a, c f z [1 A + pβa B] pη + 1 η + 1 pβ A B B 1 ηhz 1 pβa 1 B 1 B 1 B ηzh z : Φhz, zh z; z, Φr, s; t 1 pβ A B 1 B ηs + 1 pβa B 1 B For all real x and y satisfying y 1 + x, we have RΦix, y; z R{ 1 pβ A B 1 B A + pβa B] 1 ηr pη + 1 η[1. 1 B ηy + 1 pβa B 1 B 1 pβ A B A + pβa B] ηy pη + 1 η[1 1 B 1 B 1 + x 1 1 pβ A B A + pβa B] η pη + 1 η[1 1 B 1 B B A + pβa B] 1 pβa η pη + 1 η[1 1 B 1 B 63 A + pβa B] 1 ηix pη + 1 η[1 } 1 B

If we set Ω { ξ : Rξ < 1 L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1978 } B A + pβa B] 1 pβa η pη + 1 η[1, 1 B 1 B then Φix, y; z Ω for all real x, y such that y 1 + x. Furthermore, by virtue of 61, we know that Φhz, zh z; z Ω. Thus, by Lemma 1.3, we conclude that Rhz > 0 z U, which implies that the assertion of Theorem.8 holds true. Taking p 1, a c 1, A 1 α, 0 α < 1, B 1, and β 0 in Theorem.8, we revise the result of Theorem.4 in [18] and obtain the following Corollary. Corollary.9. If f Σ satisfies f z R z f 1 + η z f z z f > 1 η1 + 3α α, z then f Mα, for 0 α < 1, η 0. Theorem.10. If f Σ p satisfies anyone of the following conditions: i For B 1, { p1 B { } Lp a, c f z 1 pβa B zl p a, c f z + β + 1 AB } A B η z τ, 65 ii For B 1, A 1, 1 A1 pβ zl p a, c f z 1 + p L p a, c f z + βzl p a, c f z η z τ, 66 iii For B 1, A 1, { p L p a, c f z 1 pβ zl p a, c f z + β + 1} η z τ, 67 then f M a,c p; β; A, B, for 0 < η τ + 1 and τ 0. Proof. i For f Σ p, if B 1, we define the function Ψz by { p1 B { } Lp a, c f z Ψz z 1 pβa B zl p a, c f z + β + 1 AB } A B z U. Then Ψz is regular in U and Ψ0 0. The condition of Theorem gives us that { p1 B { } Lp a, c f z 1 pβa B zl p a, c f z + β + 1 AB } A B Ψz z η z τ. It follows that Ψz z z 0 Ψt dt t z 0 η t τ d t η τ + 1 z τ+1. This implies that Ψz z η τ + 1 z τ+1 < 1 0 < η τ + 1, τ 0.

Therefore, by the definition of Ψz, we conclude that p1 B { } Lp a, c f z 1 pβa B zl p a, c f z + β + 1 AB A B < 1. which is equivalent to { } p Lp a, c f z 1 pβ zl p a, c f z + β L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1979 + 1 AB 1 B < A B 1 B. Therefore, we conclude that f z M a,c p; β; A, B for B 1. ii For f Σ p, if B 1, A 1, we define the function Ψz by 1 A1 pβ zl p a, c f z Ψz z 1 + p L p a, c f z + βzl p a, c f z z U. Then Ψz is regular in U and Ψ0 0. The condition of Theorem gives us that 1 A1 pβ zl p a, c f z 1 + p L p a, c f z + βzl p a, c f z Ψz z η z τ. It follows that Ψz z z 0 Ψt dt t z 0 η t τ d t η τ + 1 z τ+1. This implies that Ψz z η τ + 1 z τ+1 < 1 0 < η τ + 1, τ 0. Therefore, by the definition of Ψz, we conclude that 1 A1 pβ zl p a, c f z p L p a, c f z + βzl p a, c f z + 1 < 1, which is equivalent to 1 pβ zl p a, c f z p L p a, c f z + βzl p a, c f z + 1 1 A < 1 1 A Therefore, we conclude that f z M a,c p; β; A, B for B 1, A 1. iii For f Σ p, if B 1, A 1, we define the function Ψz by p L p a, c f z Ψz z 1 pβ zl p a, c f z + β + 1 z U. A 1; z U. Then Ψz is regular in U and Ψ0 0. The condition 67 of Theorem gives us that p L p a, c f z 1 pβ zl p a, c f z + β + 1 Ψz z η z τ. It follows that Ψz z z 0 Ψt dt t z 0 η t τ d t η τ + 1 z τ+1. This implies that Ψz z η τ + 1 z τ+1 < 1 0 < η τ + 1, τ 0.

Therefore, by the definition of Ψz, we conclude that p L p a, c f z 1 pβ zl p a, c f z + β + 1 < 1, which is equivalent to p L p a, c f z 1 pβ zl p a, c f z + β L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1980 1 + z z U. 1 z Therefore, we conclude that f z M a,c p; β; A, B for B 1, A 1. Taking p 1, a c 1, A 1 α, 0 < α < 1, B 1 and β 0 in Theorem.10, we obtain the following Corollary. Corollary.11. [18] If f Σ satisfies 1 + αz f z f z η z τ, then f Mα, for 0 < α < 1, 0 < η τ + 1 and τ 0. Theorem.1. If f Σ p satisfies 1 pβ p zl p a, c f z L p a, c f z + βzl p a, c f z 1 + zl pa, c f z L p a, c f z zl pa, c f z + βzl p a, c f z L p a, c f z + βzl p a, c f z then f M a,c p; β; A, B, for 1 B < A < 1 + B. Proof. Let qz : p { } Lp a, c f z 1 pβ zl p a, c f z + β < A B 1 A, 68 z U. 69 Then the function qz is analytic in U. It follows from 69 that { 1 1 pβ zl p a, c f z } z z qz p L p a, c f z + βzl p a, c f z 1 pβ zl p a, c f z 1 p L p a, c f z + βzl p a, c f z + zl pa, c f z L p a, c f z zl pa, c f z + βzl p a, c f z L p a, c f z + βzl p a, c f z Combining 68 and 70, we find that 1 z qz < A B z U, 1 A that is z 1 A B z z U. qz 1 A An application of Lemma 1.4 to 71 yields qz 1 A : Ϝz. 7 1 A + A Bz 70 71

By noting that R 1 + zϝ z Ϝ z L.N. Ma, S.H. Li / Filomat 8:9 014, 1965 198 1981 1 A A Bz R 1 A + A Bz 1 A A B 1 A + A B 1 A + B 1 B > 0 1 B < A < 1 + B ; z U, which implies that the region ϜU is symmetric with respect to the real axis and Ϝ is convex univalent in U. Therefore, we have RϜz > Ϝ1 1 A 1 B z U. 73 Combining 69, 7 and 73, we deduce that { { }} p Lp a, c f z R 1 pβ zl p a, c f z + β < 1 A 1 B 1 B < A < 1 + B ; z U, which is equivalent to p 1 pβ { Lp a, c f z zl p a, c f z + β } 1 + Az 1 + Bz This evidently completes the proof of Theorem.1. 1 B < A < 1 + B ; z U. Putting p 1, a c 1, A 1 α, 1 < α < 1, B 1 and β 0 in Theorem.1, we obtain the following Corollary. Corollary.13. [18] If f Σ satisfies z f z 1 + z f z f z f z f z z f z < 1 α 1, then f Mα, for 1 < α < 1. Acknowledgements Authors would like to thank the referee for his valuable suggestions and comments which improve the presentation of the paper. References [1] M. K. Aouf and B. A. Frasin, Properties of some families of meromorphic multivalent functions involving certain linear operator, Filomat. 43 010 35 54. [] M. K. Aouf, A. Shamandy, A. O. Mostafa, E. A. Adwan, Integral means of certain classes of analytic functions defined by Dziok-Srivastava operator, Acta Universitatis Apulensis, 6 011 31 330. [3] T. Bulboacǎ, M. K. Aouf and R. M. El-Ashwah, Convolution properties for subclasses of meromorphic univalent functions of complex order, Filomat. 61 01 153 163. [4] P. L. Duren, Univalent functions, Grundlehren der Mathematischen Wissenschaften, Band 59, New York: Springer-Verlag, 1983. [5] J. Dziok, H. M. Srivastava, Classes of analytic functions associated with the generalized hypergeometric function, Appl. Math. Comput. 10319991 13. [6] J. Dziok, H. M. Srivastava, Certain subclasses of analytic functions associated with the generalized hypergeometric function, Integral Transform. Spec. Funct. 14003 7 18. [7] I. S. Jack, Functions starlike and convex of order α, J. London Math. Soc. 3 1971 469 474.

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